$\Delta H$ for $\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})$ is $176 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $1240 \mathrm{~K}.$ The $\Delta U$ for the reaction is equal to

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Easy

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Application +4 | -1

\Delta H

for

\mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})

176 \mathrm{~kJ} \mathrm{~mol}^{-1}

1240 \mathrm{~K}.

The

\Delta U

for the reaction is equal to

View answer and explanation

Correct option: B

165.6 \mathrm{~kJ}

The

\Delta U

for the reaction is 165.6 kJ, calculated using the formula

\Delta H = \Delta U + \Delta n_g RT

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{T_i}

{T_f}

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

300 \mathrm{~K}

\Delta G